# The velocity $v$ of a particle moving along a straight line when at a distance $x$ from the origin is given by $a+by^{2}=x^{2}$ where $a$ and $b$ are constants. Then the acceleration is

$\begin{array}{1 1}(1)\frac{b}{x}&(2)\frac{a}{x}\\(3)\frac{x}{b}&(4)\frac{x}{a}\end {array}$

$a+bv^2=x^2$
$b.2v. \large\frac{dv}{dt}$$=2x. \large\frac{dx}{dt} 2b.v. \large\frac{dv}{dt}$$=2x.v$
Acceleration $\large\frac{dv}{dt} =\frac{x}{b}$
Hence 3 is the correct answer.